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Suppose we have two integers $b$ and $c$.

If $\gcd(b, c) = 1$, then the pair of integers is relatively prime.

$\gcd(b, c) = 1$ can also be expressed as a linear combination $bx_0 + cy_0 =1$, where $x_0$ and $y_0$ are integers chosen such that $bx_0 + cy_0$ is the smallest positive element in the set $ \{bx + cy\} $, where $x$ and $y$ are integers.

From the above, we can conclude that all pairs of consecutive integers are relatively prime pairs considering:

$(1)(x+1) + (-1)(x) = 1 \implies \gcd(x, x+1) = 1$ for an integer $x$

Now, consider the relatively prime pair of numbers $37$ and $39$. Because $\gcd(37, 39) = 1$ we know that there must be some $x_0$ and some $y_0$ for which $37x_0 + 39y_0 = 1$. That means there are two consecutive integers represented by $37x_0$ and $39y_0$. (We also know that a pair of consecutive integers must correspond to every pair of relatively prime numbers that is not consecutive because of equality of the linear combination to 1).

I have found $x_0 = 19, y_0 = 18$ representing the pair of consecutive numbers $\gcd(703, 702)$ that is related to $\gcd(37, 39)$.

I want to better understand the relationship between pairs of relatively prime numbers that are not consecutive numbers and their relatively prime numbers that are consecutive numbers. For example, is there a unique pair of consecutive numbers for every relatively prime set of nonconsecutive numbers?

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For your latter question, no.

$$(6)+(-5) = (2){\color{red}{(3)}}+(-1){\color{red}{(5)}}=1$$ $$(-9)+(10) = (-3){\color{red}{(3)}}+(2){\color{red}{(5)}}=1$$

For $(a,b)=1$, then there exist integers $m,n$ with $ma+nb=1$.

By adding and subtracting $ab$, we get a new solution

$$ma+nb +ab - ab = 1 \iff (m+b)a + (n-a)b = 1$$

You can continue this indefinitely.

And so one question you may ask is:

How often can the sequence $\{(|(m+kb)a|, |(n-ka)b|)\}_{k\in\mathbb Z}$ be a consecutive pair?

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  • $\begingroup$ Would there be a way to know exactly how many pairs of consecutive numbers are associated with any pair of nonconsecutive numbers? $\endgroup$ – Richard K Yu Dec 1 '19 at 1:39
  • $\begingroup$ Hopefully my edit helps you some $\endgroup$ – David P Dec 1 '19 at 2:48

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